Question 24555: what is the center of the circle x^2 + 4x + y^2 + 2y - 9 = 0? Found 3 solutions by kev82, askmemath, Earlsdon:Answer by kev82(151) (Show Source):
You can put this solution on YOUR website! Hi,
The general equation of a circle as I'm sure you know is where (a,b) is the centerpoint and r is the radius. We can make your equation look like this one by completing the square.
Firstly we're looking for the value of a such that You can either just guess that or expand and compare coefficients. But So
Doing the same for y, we get Finally substituting into the equation we get
So, in the standard form for a circle we get
So we can read off from this that the circle is centered at (-2,-1) with a radius of
Hope that helps,
Kev
You can put this solution on YOUR website! Equation of a circle is given by where (a,b) is the center of the circle and r is the radius.
Therefore we need to write x^2 + 4x + y^2 + 2y - 9 = 0 in form above.
Since I have all i need to do is add 4 and i can re-write it as However if I am adding 4 then I also need to subtract 4 from the equation
So i get
Similarly for I can add 1 and re-write as and similarly subtract 1
Adding 14 on both sides
so center of circle becomes (-2,-1) and radius becomes
You can put this solution on YOUR website! Find the centre of the circle:
First get your equation into the standard form for a circle. You can do this by "completing the square" in the x-terms and in the y-terms. Complete the square in the x- and y-terms by adding the square of half the x-, y-coefficients to both sides of the equation. Factor and Simplify. Add 9 to both sides. Now compare this with the standard form for a circle with centre at (h, k).
You can see that the centre of the circle is at (-2, -1)
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